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Step-by-Step Solution
Step 1: Identify the Physical Situation
The rod of mass M = 2\text{ kg} and length L = 0.6\text{ m} is pivoted at its lower end and is initially vertical. It is allowed to rotate freely under gravity until its free (top) end passes through the lowest position (i.e., it rotates by 180° from pointing up to pointing down).
Step 2: Determine the Loss in Potential Energy
The center of mass of the rod is initially at a height of L/2 = 0.6/2 = 0.3\text{ m} above the pivot. After rotating by 180°, the center of mass is at 0.3\text{ m} below the pivot. Thus, the change in height of the center of mass is \Delta h = 0.6\text{ m} . The corresponding loss in gravitational potential energy ( \Delta U ) is:
\Delta U = Mg\,\Delta h = 2 \times 10 \times 0.6 = 12\text{ J}.
Step 3: Relate the Potential Energy Change to Rotational Kinetic Energy
When the rod reaches its lowest position, all of this lost potential energy converts into rotational kinetic energy ( K ). Mathematically,
\Delta U = K = \frac{1}{2} I \omega^2,
where I is the moment of inertia of the rod about its lower end, and \omega is its angular speed at that instant.
Step 4: Calculate the Moment of Inertia of the Rod
For a uniform rod of mass M and length L pivoted about one end, the moment of inertia is:
I = \frac{1}{3} M L^2.
Hence,
I = \frac{1}{3} \times 2 \times (0.6)^2
= \frac{2}{3} \times 0.36
= 0.24\text{ kg·m}^2.
Step 5: Determine the Angular Speed
Using the energy relation \frac{1}{2} I \omega^2 = 12 , we get:
\frac{1}{2} \times 0.24 \times \omega^2 = 12 \\
0.12 \,\omega^2 = 12 \\
\omega^2 = 100 \\
\omega = 10\text{ rad/s}.
Step 6: Find the Linear Speed of the Free End
The linear speed v of the tip of the rod (free end) is related to the angular speed by v = \omega \times r , where r = L = 0.6\text{ m} . Therefore,
v = 10 \times 0.6 = 6\text{ m/s}.
Step 7: Final Answer
The speed of the free end of the rod when it passes through its lowest position is 6\text{ m/s} .
